Centralizing Traces and Lie Triple Isomorphisms on Triangular Algebras

0.26 MB·English

by  Xinfeng Liang

CENTRALIZING TRACES AND LIE TRIPLE ISOMORPHISMS
ON TRIANGULAR ALGEBRAS
3
1 XINFENGLIANG,ZHANKUIXIAOANDFENGWEI
0
2
Abstract. LetT beatriangularalgebraoveracommutativeringRandZ(T)
n be the center of T. Suppose that q:T ×T −→ T is an R-bilinear mapping
a andthatTq: :T −→T isatraceofq. WedescribetheformofTq satisfying
J the condition [Tq(T),T]∈Z(T)for all T ∈T. The question of when Tq has
0 the proper form will be addressed. Using the aforementioned trace function,
1 we establish sufficient conditions for each Lie triple isomorphism on T to be
almost standard. As applications we characterize Lie triple isomorphisms of

] triangular matrix algebras and nest algebras. Some further research topics
A
relatedtocurrentworkareproposedattheendofthisarticle.
R
.
h
t
a
m 1. Introduction
[ Let R be a commutative ring with identity, A be a unital algebra over R and

1 Z(A) be the center of A. Let us denote the commutator or the Lie product of the
v elements a,b∈A by [a,b]=ab−ba. Recall that an R-linear mapping f:A−→A
3 is said to be semi-centralizing if either [f(a),a] ∈ Z(A) or f(a)a+af(a) ∈ Z(A)
4
for all a ∈ A. Further, the mapping f is said to be centralizing if [f(a),a] ∈ Z(A)
0
for all a ∈ A. The mapping f is said to be skew-centralizing if f(a)a+ af(a) ∈
2
. Z(A) for all a ∈ A. In particular, the mapping f is said to be commuting if
1
[f(a),a] = 0 for all a ∈ A. The mapping f is said to be skew-commuting if f(a)a+
0
af(a) = 0 for all a ∈ A. When we investigate the above-mentioned mappings, the
3
1 principal task is to describe their forms. This is demonstrated by various works,
: see [8, 10, 11, 12, 13, 15, 20, 21, 28, 29, 33, 36, 41, 44, 45, 49, 50]. We encourage
v
thereadertoreadthewell-writtensurveypaper[13],inwhichtheauthorpresented
i
X
thedevelopmentofthetheoryofsemi-centralizingmappingsandtheirapplications
r in details.
a
Let R be a commutative ring with identity, A be a unital algebra over R and
Z(A) be the center of A. Recall that an R-linear mapping f : A −→ A is said
to be centralizing if [f(a),a] ∈ Z(A) for all a ∈ A. Let n be a positive integer
and q: An −→ A be an n-linear mapping. The mapping T : A −→ A defined by
q
T (a) =q(a,a,··· ,a) is called a trace of q. We say that a centralizing trace T is
q q
proper if it can be written as
Tq(a)=zan+µ1(a)an−1+···+µn−1(a)a+µn(a)
2000 Mathematics Subject Classification. 47L35,15A78,16W25.
Keywordsandphrases. Centralizingtrace,Lietripleisomorphism,commutingtrace,triangu-
laralgebra,nestalgebra.
The work of the second author is supported by the Mathematical Tianyuan Fundamental of
NSFC(GrantNo. 11226068).
1
2 XINFENGLIANG,ZHANKUIXIAOANDFENGWEI
for all a ∈ A, where z ∈ Z(A) and µ (1 ≤ i ≤ n) is a mapping from A into
i
Z(A) and every µ (1≤i≤n) is in fact a trace of an i-linear mapping q from Ai
i i
into Z(A). Let n = 1 and f: A −→ A be an R-linear mapping. In this case, an
arbitrary trace T of f exactly equals to itself. Moreover, if a centralizing trace T
f f
of f is proper, then it has the form
T (a)≡za modZ(A), ∀a∈A,
f
where z ∈Z(A). Let us see the case of n=2. Suppose that g: A×A−→A is an
R-bilinear mapping. If a centralizing trace T of g is proper, then it is of the form
g
T (a)≡za2+µ(a)a modZ(A), ∀a∈A,
g
where z ∈ Z(A) and µ is an R-linear mapping from A into Z(A). It was Breˇsar
who initiated the study of commuting traces and centralizing traces of bilinear
mappings in his series of works [10, 11, 12, 13, 15], where he investigated the
structure of commuting traces and centralizing traces of (bi-)linear mappings on
prime rings. It has turned out that in certain rings, in particular, prime rings
of characteristic different from 2 and 3, every centralizing trace of a biadditive
mapping is commuting. Moreover, every centralizing mapping of a prime ring of
characteristic not 2 is of the proper form and is actually commuting. Lee et al
further generalized Breˇsar’s results by showing that each commuting trace of an
arbitrary multilinear mapping on a prime ring also has the proper form [28].
Cheungin[21]studiedcommutingmappingsoftriangularalgebras(e.g.,ofupper
triangularmatrixalgebrasandnestalgebras). Hedeterminedtheclassoftriangular
algebrasforwhicheverycommutingmappingisproper. XiaoandWei[49]extended
Cheung’s result to the generalized matrix algebra case. Motivated by the results
of Breˇsar and Cheung, Benkoviˇc and Eremita [8] considered commuting traces of
bilinearmappingsonatriangularalgebra[AM]. Theygaveconditionsunderwhich
O B
everycommutingtraceofatriangularalgebra[A M]isproper. Inviewoftheabove
O B
works,itisnaturalandnecessarytocharacterizecentralizingtracesof(multi-)linear
mappings on triangularalgebras. One of the main aims ofthis article is to provide
a sufficient condition for each centralizing trace of an arbitrary bilinear mapping
on a triangular algebra [AM] to be proper.
O B
Another important purpose of this article is to address the Lie triple isomor-
phisms problem of triangular algebras. At his 1961 AMS Hour Talk, Herstein
proposed many problems concerning the structure of Jordan and Lie mappings in
associativesimpleandprimerings[26]. TherenownedHerstein’sLie-typemappings
research program was formulated since then. The involved Lie mappings mainly
include Lie isomorphisms, Lie triple isomorphisms, Lie derivations and Lie triple
derivations et al. Given a commutative ring R with identity and two associative
R-algebras A and B, one define a Lie triple isomorphism from A into B to be an
R-linear bijective mapping l satisfying the condition
l([[a,b],c])=[[l(a),l(b)],l(c)] ∀a,b,c∈A.
For example, an isomorphism or a negative of an anti-isomorphism of one algebra
onto another is also a Lie isomorphism. Furthermore, every Lie isomorphism and
every Jordan isomorphism are Lie triple isomorphisms. One can ask whether the
converseistrueinsomespecialcases. Thatis,doeseveryLietripleisomorphismbe-
tween certain associative algebras arise from isomorphisms and anti-isomorphisms
CENTRALIZING TRACES AND LIE TRIPLE ISOMORPHISMS 3
inthe sense ofmodulo mappingswhoserangeiscentral? Recallthata Lieisomor-
phism l: A−→B is standard if
l=m+n, (♣)
where m is an isomorphism or the negative of an anti-isomorphism from A onto B
and n:A−→Z(B) is an R-linear mapping annihilating all commutators. We say
that a Lie triple isomorphism l: A−→B is standard if
l=±m+n, (♠)
where m is an isomorphism or an anti-isomorphism from A onto B and n : A −→
Z(B) is an R-linear mapping annihilating all second commutators.
The resolution of Herstein’s Lie isomorphisms problem in matrix algebra back-
ground has been well-known for a long time. Hua [27] proved that every Lie au-
tomorphism of the full matrix algebra M (D)(n ≥ 3) over a division ring D is of
n
the standard form (♣). This result was extended to the nonlinear case by Dolinar
[24] and was further refined by Sˇemrl [44]. Dokovi´c [23] showed that every Lie
automorphismof upper triangular matrix algebrasT (R) over a commutative ring
n
R without nontrivial idempotents has the standard form as well. Marcoux and
Sourour [33] classified the linear mappings preserving commutativity in both di-
rections (i.e., [x,y] = 0 if and only if [f(x),f(y)] = 0) on upper triangular matrix
algebras T (F) over a field F. Such a mapping is either the sum of an algebra
n
automorphism of T (F) (which is inner) and a mapping into the center FI, or
n
the sum of the negative of an algebra anti-automorphism and a mapping into
the center FI. The classification of the Lie automorphisms of T (F) is obtained
n
as a consequence. Benkoviˇc and Eremita [8] applied the theory of commuting
traces to study the Lie isomorphisms on a triangular algebra. They provided suf-
ficient conditions under which every commuting trace of triangular algebra [AM]
O B
is proper. It also turns out that under some mild assumptions, each Lie isomor-
phismof [AM]has the standardform(♣). Caldero´nMart´ınandMart´ınGonz´alez
O B
observed that every Lie triple isomorphism of the full matrix algebra M (C) over
n
the complex field C is of the standard form (♠) [18]. Simultaneously, Lie triple
isomorphismsbetween rings and between (non-)self-adjoint operatoralgebras have
received a fair amount of attentions. The involved rings and operator algebras
include (semi-)prime rings, the algebra of bounded linear operators, C∗-algebras,
von Neumann algebras, H∗-algebras, nest algebras, reflexive algebras and so on,
see [16, 17, 18, 19, 31, 32, 34, 35, 37, 38, 39, 42, 43, 44, 45, 46, 51, 52].
This is the second paper in a series of three that we are planning on this topic.
The first paper was dedicated to studying, in more details, commuting traces and
Lie isomorphisms on generalized matrix algebras [50]. This article is organized as
following. Section 2 contains the definition of triangular algebra and some clas-
sical examples. In Section 3 we provide sufficient conditions for each centralizing
trace of arbitrary bilinear mappings on a triangular algebra [A M] to be proper
O B
(Theorem 3.4). And then we apply this result to describe the centralizing traces
of bilinear mappings on certain classical triangular algebras. In Section 4 we will
give sufficient conditions under which every Lie triple isomorphism from a trian-
gular algebra into another one has the almost standard form (Theorem 4.4). As
corollaries of Theorem 4.4, characterizations of Lie triple isomorphisms on several
kinds of triangular algebrasare obtained. The last section contains some potential
future researchtopics related to our current work.
4 XINFENGLIANG,ZHANKUIXIAOANDFENGWEI
2. Preliminaries
LetRbe acommutativeringwithidentity. LetAandB be unitalalgebrasover
R. Recall that an (A,B)-bimodule M is loyal if aMb = 0 implies that a = 0 or
b = 0 for any a ∈ A,b ∈ B. Clearly, each loyal (A,B)-bimodule M is faithful as a
left A-module and also as a right B-module.
Let A,B be unital associative algebras over R and M be a unital (A,B)-
bimodule, which is faithful as a left A-module and also as a right B-module. We
denote the triangular algebra consisting of A,B and M by
A M
T = .
(cid:20) 0 B (cid:21)
Then T is an associative and noncommutative R-algebra. The center Z(T) of T
is (see [21, Proposition 3])
a 0
Z(T)= am=mb, ∀ m∈M .
(cid:26)(cid:20) 0 b (cid:21) (cid:27)
Let us define two natural R-linear projections π :T →A and π :T →B by
A B
a m a m
π : 7−→a and π : 7−→b.
A (cid:20) 0 b (cid:21) B (cid:20) 0 b (cid:21)
It is easy to see that π (Z(T)) is a subalgebra of Z(A) and that π (Z(T)) is
A B
a subalgebra of Z(B). Furthermore, there exists a unique algebraic isomorphism
τ: π (Z(T)) −→ π (Z(T)) such that am = mτ(a) for all a ∈ π (Z(T)) and for
A B A
all m∈M.
Let 1 (resp. 1′) be the identity of the algebra A (resp. B), and let I be the
identity of the triangular algebra T. We will use the following notations:
1 0 0 0
P = , Q=I−P =
(cid:20) 0 0 (cid:21) (cid:20) 0 1′ (cid:21)
and
T =PTP, T =PTQ, T =QTQ.
11 12 22
Thus the triangular algebra T can be written as
T =PTP +PTQ+QTQ=T +T +T .
11 12 22
T andT aresubalgebrasofT whichareisomorphictoAandB,respectively. T
11 22 12
isa(T ,T )-bimodulewhichisisomorphictothe(A,B)-bimoduleM. Itshouldbe
11 22
remarkedthatπ (Z(T))andπ (Z(T))areisomorphictoPZ(T)P andQZ(T)Q,
A B
respectively. Thenthere isanalgebraisomorphismτ: PZ(T)P −→QZ(T)Qsuch
that am=mτ(a) for all m∈PTQ.
Let us list some classical examples of triangular algebras and matrix algebras
which will be revisited in the sequel (Section 3, Section 4 and Section5). Since
these examples have already been presented in many papers, we just state their
titles without any introduction. We refer the reader to [8, 29, 49] for more details.
(a) Upper and lower triangular matrix algebras;
(b) Block upper and lower triangular matrix algebras;
(c) Hilbert space nest algebras;
(d) Full matrix algebras;
(e) Inflated algebras.
CENTRALIZING TRACES AND LIE TRIPLE ISOMORPHISMS 5
3. Centralizing Traces of Triangular Algebras
Inthissectionwewillestablishsufficientconditionsforeachcommutingtraceof
arbitrary bilinear mappings on a triangular algebra [AM] to be proper (Theorem
O B
3.4). Consequently,we areableto describe centralizingtracesofbilinearmappings
on upper triangular matrix algebras and nest algebras. The most important fact
is that Theorem 3.4 will be used to characterize Lie triple isomorphisms from a
triangular algebra into another in Section 4.
We now list some basic facts related to triangular algebras, which can be found
in [8, Section 2].
Lemma 3.1. Let M be a loyal (A,B)-bimodule and let f,g: M → A be arbitrary
mappings. Suppose f(m)n+g(n)m=0 for all m,n∈M. If B is noncommutative,
then f =g =0.
Lemma 3.2. Let T = [A M] be a triangular algebra with a loyal (A,B)-bimodule
O B
M, λ∈π (Z(T)) and b∈B be a nonzero element. If λb=0, then λ=0
B
Lemma 3.3. Let T = [A M] be a triangular algebra with a loyal (A,B)-bimodule
O B
M. Then the center Z(T) of T is a domain.
We are in position to state the main theorem of this section.
Theorem 3.4. Let T = [AM] be a 2-torsion free triangular algebra over the
O B
commutative ring R and q: T ×T −→T be an R-bilinear mapping. If
(1) each commuting linear mapping on A or B is proper,
(2) π (Z(T))=Z(A)6=A and π (Z(T))=Z(B)6=B,
A B
(3) M is loyal,
then every centralizing trace T :T −→T of q is proper.
q
For convenience, let us write A = A, A = B and A = M. We denote the
1 2 3
unityofA by1andtheunityofA by1′. SupposethatT isanarbitrarytraceof
1 2 q
the R-bilinear mapping q. Then there exist bilinear mappings f : A ×A →A ,
ij i j 1
g : A ×A →A and h : A ×A →A (16i6j 63) such that
ij i j 2 ij i j 3
a a F(a ,a ,a ) H(a ,a ,a )
T : 1 3 7→ 1 2 3 1 2 3 ,
q (cid:20) a2 (cid:21) (cid:20) G(a1,a2,a3) (cid:21)
where
F(a ,a ,a )= f (a ,a ),
1 2 3 ij i j
16Xi6j63
G(a ,a ,a )= g (a ,a ),
1 2 3 ij i j
16Xi6j63
H(a ,a ,a )= h (a ,a ).
1 2 3 ij i j
16Xi6j63
Since T is centralizing, we have
q
F H a a [F,a ] Fa +Ha −a H −a G
, 1 3 = 1 3 2 1 3 ∈Z(T).
(cid:20)(cid:20) G (cid:21) (cid:20) a2 (cid:21)(cid:21) (cid:20) [G,a2] (cid:21)
(3.1)
Now we divide the proof of Theorem3.4 into a series of lemmas for comfortable
reading.
6 XINFENGLIANG,ZHANKUIXIAOANDFENGWEI
Lemma 3.5. Let K : A ×A → A (resp. K : A ×A → A ) be an R-bilinear
2 2 3 1 1 3
mapping. If K(x,x)x = 0 (resp. xK(x,x) = 0 ) for all x ∈ A (resp. for all
2
x∈A ), then K(x,x)=0.
1
Proof. Setting x = 1′, we obtain that K(1′,1′) = 0. Replacing x by x + 1′ in
K(x,x)x=0, we get
K(x,x)=−(K(1′,x)+K(x,1′))(1′+x). (3.2)
Substituting x−1′ for x in K(x,x)x=0, we arrive at
K(x,x)=(K(1′,x)+K(x,1′))(1′−x). (3.3)
Combining the above two relations gives K(1′,x)+K(x,1′) = 0. Thus K(x,x) =
0. (cid:3)
Lemma 3.6. H(a ,a ,a )=h (a ,a )+h (a ,a )+h (a ,a ).
1 2 3 13 1 3 23 2 3 33 3 3
Proof. It follows from the matrix relation (3.1) that
Fa +Ha −a H −a G=0. (3.4)
3 2 1 3
Let us take a =0 and a =0 into (3.4). Then (3.1) implies that
1 2
f (a ,a )a =a g (a ,a ) (3.5)
33 3 3 3 3 33 3 3
for all a ∈ A . Let us choose a = 0 and a = 0 in (3.4). Then 0 = Ha =
3 3 1 3 2
h (a ,a )a for all a ∈ A . In view of Lemma 3.5, we have h (a ,a ) = 0.
22 2 2 2 2 2 22 2 2
Similarly, putting a =0 and a =0 in (3.4) yields h (a ,a )=0 for all a ∈A .
2 3 11 1 1 1 1
Furthermore, setting a =0 in (3.4), we see that
3
(h (a ,a )a −a h (a ,a ))=0
12 1 2 2 1 12 1 2
for all a ∈A ,a ∈A . Replacing a by −a in the above relationand comparing
1 1 2 2 1 1
the obtained two relations gives a h (a ,a )=0 for all a ∈A ,a ∈A . In par-
1 12 1 2 1 1 2 2
ticular,h (1,a )=0foralla ∈A . Substitutinga +1fora ina h (a ,a )=0
12 2 2 2 1 1 1 12 1 2
leads to h (a ,a )=0 for all a ∈A ,a ∈A . Therefore
12 1 2 1 1 2 2
H(a ,a ,a )=h (a ,a )+h (a ,a )+h (a ,a )
1 2 3 13 1 3 23 2 3 33 3 3
for all a ∈A ,a ∈A ,a ∈A . (cid:3)
1 1 2 2 3 3
Lemma 3.7. With notations as above, we have
(1) a 7→f (a ,a ) is a commuting trace,
1 11 1 1
a 7→f (a ,a ) is a commuting linear mapping for each a ∈A ,
1 13 1 3 3 3
a 7→g (a ,a ) is a commuting trace,
2 22 2 2
a 7→g (a ,a ) is a commuting linear mapping for each a ∈A ,
2 23 2 3 3 3
(2) [g (a ,a ),a ]=τ([f (a ,a ),a ])∈Z(A ),
11 1 1 2 12 1 2 1 2
[g (a ,a ),a ]=τ([f (a ,a ),a ])∈Z(A ),
12 1 2 2 22 2 2 1 2
[g (a ,a ),a ]=τ([f (a ,a ),a ])∈Z(A ),
13 1 3 2 23 2 3 1 2
(3) f (a ,a )∈Z(A ) and g (a ,a )∈Z(A ).
33 3 3 1 33 3 3 2
Proof. By the relation (3.1) we know that
τ([F,a ])=[G,a ]. (3.6)
1 2
Let us take a =0 in (3.6). Then
1
[g (a ,a )+g (a ,a )+g (a ,a ),a ]=0 (3.7)
22 2 2 23 2 3 33 3 3 2
CENTRALIZING TRACES AND LIE TRIPLE ISOMORPHISMS 7
for all a ∈A ,a ∈A . Replacing a by −a in (3.7) we get
2 2 3 3 3 3
[g (a ,a )+g (a ,a ),a ]=0 (3.8)
22 2 2 33 3 3 2
for all a ∈A ,a ∈A . Putting a =0 in (3.7) and combining (3.7) and (3.8), we
2 2 3 3 3
obtain
[g (a ,a ),a ]=0, [g (a ,a ),a ]=0, [g (a ,a ),a ]=0
22 2 2 2 23 2 3 2 33 3 3 2
for all a ∈A ,a ∈A . In a similar way, we have
2 2 3 3
[f (a ,a ),a ]=0, [f (a ,a ),a ]=0, [f (a ,a ),a ]=0.
11 1 1 1 33 3 3 1 13 1 3 1
Setting a =0 in (3.6), we arrive at
3
τ([f (a ,a )+f (a ,a ),a ])=[g (a ,a )+g (a ,a ),a ] (3.9)
12 1 2 22 2 2 1 11 1 1 12 1 2 2
for all a ∈ A ,a ∈ A . Replacing a by −a in (3.9) and then comparing the
1 1 2 2 1 1
obtained relation with (3.9), we get
τ([f (a ,a ),a ])=[g (a ,a ),a ]∈Z(A ) (3.10)
22 2 2 1 12 1 2 2 2
and
τ([f (a ,a ),a ])=[g (a ,a ),a ]∈Z(A ) (3.11)
12 1 2 1 11 1 1 2 2
for all a ∈A ,a ∈A . In view of (3.6),(3.10),(3.11)we conclude
1 1 2 2
τ([f (a ,a ),a ])=[g (a ,a ),a ]∈Z(A )
23 2 3 1 13 1 3 2 2
for all a ∈A ,a ∈A ,a ∈A . (cid:3)
1 1 2 2 3 3
Lemma 3.8. There exist alinear mapping ξ :A →Z(A ) anda bilinear mapping
3 2
η :A ×A →Z(A ) such that g (a ,a )=ξ(a )a +η(a ,a ).
2 3 2 23 2 3 3 2 2 3
Proof. Since a 7→ g (a ,a ) is a commuting linear mapping for each a ∈ A ,
2 23 2 3 3 3
thenbythehypothesis(1)thereexistmappingsξ :A →Z(A )andη :A ×A →
3 2 2 3
Z(A ) such that
2
g (a ,a )=ξ(a )a +η(a ,a ),
23 2 3 3 2 2 3
where η is R-linear in the first argument. Let us show that ξ is R-linear and η is
R-bilinear. Clearly,
g (a ,a +b )=ξ(a +b )a +η(a ,a +b )
23 2 3 3 3 3 2 2 3 3
g (a ,a )+g (a ,b )=ξ(a )a +η(a ,a )+ξ(b )a +η(a ,b )
23 2 3 23 2 3 3 2 2 3 3 2 2 3
for all a ∈A ,a ,b ∈A . So
2 2 3 3 3
ξ(a +b )−ξ(a )−ξ(b ) a +η(a ,a +b )−η(a ,a )−η(a ,b )=0
3 3 3 3 2 2 3 3 2 3 2 3
(cid:0) (cid:1)
for all a ∈ A ,a ,b ∈ A . Note that ξ and η map into Z(A ). Hence (ξ(a +
2 2 3 3 3 2 3
b ) −ξ(a )− ξ(b ))[a ,b ] = 0 for all a ,b ∈ A , and a ,b ∈ A . Note that
3 3 3 2 2 2 2 2 3 3 3
A is noncommutative. Applying Lemma 3.2 yields that ξ is R-linear mapping.
2
Consequently, η is R-linear in the second argument. (cid:3)
Lemma 3.9. f (a ,a )∈Z(A ) and g (a ,a )∈Z(A ).
23 2 3 1 13 1 3 2
Proof. By Lemma 3.7 it is enough to prove f (a ,a )∈Z(A ). Setting a =0 in
23 2 3 1 1
(3.4) and using (3.5), we obtain
f (a ,a )+f (a ,a ) a + h (a ,a )+h (a ,a ) a
22 2 2 23 2 3 3 33 3 3 23 2 3 2
(3.12)
(cid:0) (cid:1) (cid:0) (cid:1)
−a g (a ,a )+g (a ,a ) =0
3 22 2 2 23 2 3
(cid:0) (cid:1)
8 XINFENGLIANG,ZHANKUIXIAOANDFENGWEI
for all a ∈ A ,a ∈ A . Replacing a by −a in the equation (3.12) and then
2 2 3 3 2 2
comparing with it, we get
h (a ,a )a =a g (a ,a )−f (a ,a )a (3.13)
23 2 3 2 3 22 2 2 22 2 2 3
and
h (a ,a )a =a g (a ,a )−f (a ,a )a (3.14)
33 3 3 2 3 23 2 3 23 2 3 3
for all a ∈ A ,a ∈ A . Note that [g (a ,a ),a ] = 0 for all a ∈ A ,a ∈ A .
2 2 3 3 23 2 3 2 2 2 3 3
Replacing a by a +1′ in [g (a ,a ),a ] = 0 gives g (1′,a ) ∈ Z(A ). On the
2 2 23 2 3 2 23 3 2
other hand, Lemma 3.7 shows that f (1′,a ) ∈ Z(A ) for all a ∈ A . Taking
23 3 1 3 3
a =1′ in (3.14) we have
2
h (a ,a )=a α(a ), (3.15)
33 3 3 3 3
where α(a ) = g (1′,a )−τ(f (1′,a )) ∈ Z(A ). It follows from (3.14), (3.15)
3 23 3 23 3 2
and Lemma 3.8 that
a (α(a )−ξ(a ))a = τ−1(η(a ,a ))−f (a ,a ) a . (3.16)
3 3 3 2 2 3 23 2 3 3
We denote Y(a )=α(a )−ξ(a ), X(cid:0)(a ,a )=τ−1(η(a ,a ))−f(cid:1) (a ,a ). Taking
3 3 3 2 3 2 3 23 2 3
a =1′ into (3.16), we see that (τ−1(Y(a ))−X(1′,a ))a =0 for all a ∈A .
2 3 3 3 3 3
We claim that
Y(a )=τ(X(1′,a )) (3.17)
3 3
for all a ∈ A . In fact, replacing a by m+n in (τ−1(Y(a ))−X(1′,a ))a = 0,
3 3 3 3 3 3
we get
(τ−1(Y(m))−X(1′,m))n+(τ−1(Y(n))−X(1′,n))m=0
for all m,n∈A . Applying Lemma 3.1 yields Y(m)=τ(X(1′,m)) for all m∈A .
3 3
Thus our claim follows.
Now let us rewrite the relation (3.16) as
a τ(X(1′,a ))a =X(a ,a )a (3.18)
3 3 2 2 3 3
for all a ∈A . Replacing a by m+n in (3.18), we obtain
3 3 3
mτ(X(1′,n))a +nτ(X(1′,m))a =X(a ,n)m+X(a ,m)n (3.19)
2 2 2 2
for all a ∈ A , m,n ∈ A . Replacing n by a n in (19) and then subtracting the
2 2 3 1
left multiplication of (3.19) by a , we arrive at
1
mτ(X(1′,a n))a −a mτ(X(1′,n))a
1 2 1 2
(3.20)
=X(a ,m)a n+X(a ,a n)m−a X(a ,m)n−a X(a ,n)m
2 1 2 1 1 2 1 2
for all a ∈A , a ∈A and m,n ∈ A . Taking m =n in (3.20) and using (3.18),
1 1 2 2 3
we have
mτ(X(1′,a m))a =X(a ,a m)m+[X(a ,m),a ]m (3.21)
1 2 2 1 2 1
for all a ∈ A ,a ∈ A ,m ∈ A . Left multiplying a in (3.21) and considering
1 1 2 2 3 1
(3.18), we get [X(a ,a m),a ]m=a [X(a ,m),a ]m. That is,
2 1 1 1 2 1
([X(a ,a m),a ]−a [X(a ,m),a ])m=0
2 1 1 1 2 1
for all a ∈ A , a ∈ A and m ∈ A . Let us write P(m) = [X(a ,a m),a ]−
1 1 2 2 3 2 1 1
a [X(a ,m),a ] for some fixed a ,a . Then P: A → A is an R-linear mapping
1 2 1 1 2 3 1
for each a ∈ A ,a ∈ A , and P(m)m = 0. A linearization of P(m)m = 0 shows
1 1 2 2
P(m)n + P(n)m = 0 for all m,n ∈ A . In view of Lemma 3.1 we know that
3
P(m)=0. So
[X(a ,a m),a ]=a [X(a ,m),a ]
2 1 1 1 2 1
CENTRALIZING TRACES AND LIE TRIPLE ISOMORPHISMS 9
for all a ∈ A ,a ∈ A ,m ∈ A . Picking b ∈ A such that [a ,b ] 6= 0, and
1 1 2 2 3 1 1 1 1
then commuting with b , we get [a ,b ][X(a ,m),a ] = 0 since [X(a ,m),a ] =
1 1 1 2 1 2 1
[a ,f (a ,m)] ∈Z(A ) by Lemma 3.7. Thus Lemma 3.2 implies [X(a ,m),a ]=
1 23 2 1 2 1
[a ,f (a ,a )]=0 and this completes the proof of the lemma. (cid:3)
1 23 2 3
Lemma 3.10. With notations as above, we have
(1) f (a ,a )∈Z(A ) and g (a ,a )∈Z(A );
22 2 2 1 11 1 1 2
(2) a 7→f (a ,a ) is a commuting linear mapping for each a ∈A ,
1 12 1 2 2 2
a 7→g (a ,a ) is a commuting linear mapping for each a ∈A .
2 12 1 2 1 1
Proof. Taking a =0 in (3.4) and using (3.5), we get
2
(f (a ,a )+f (a ,a ))a −a (g (a ,a )+g (a ,a ))
11 1 1 13 1 3 3 3 11 1 1 13 1 3
(3.22)
−a (h (a ,a )+h (a ,a ))=0
1 13 1 3 33 3 3
for all a ∈ A ,a ∈ A . Note that R is 2-torsion free ring. Substituting −a for
1 1 3 3 1
a in (3.22), we obtain
1
a h (a ,a )=f (a ,a )a −a g (a ,a ) (3.23)
1 13 1 3 11 1 1 3 3 11 1 1
for all a ∈A ,a ∈A . Combining (3.22) with (3.23) gives
1 1 3 3
a h (a ,a )=f (a ,a )a −a g (a ,a ) (3.24)
1 33 3 3 13 1 3 3 3 13 1 3
for all a ∈ A ,a ∈ A . On the other hand, replacing a by a a in (3.13) and
1 1 3 3 3 1 3
subtracting the left multiplication of (3.13) by a we get
1
(a h (a ,a )−h (a ,a a ))a =[f (a ,a ),a ]a (3.25)
1 23 2 3 23 2 1 3 2 22 2 2 1 3
for all a ∈ A ,a ∈ A ,a ∈ A . Replacing a by a a in (3.13) and subtracting
1 1 2 2 3 3 3 3 2
the right multiplication of (3.11) by a we get h (a ,a a )a = h (a ,a )a a .
2 23 2 3 2 2 23 2 3 2 2
Let us set K(x,y) = h (x,a y)−h (x,a )y, where x,y ∈ A . It is easy to see
23 3 23 3 2
that K(x,y): A ×A → A is an R-bilinear mapping, and K(a ,a )a = 0. It
2 2 3 2 2 2
follows from Lemma 3.5 that
h (a ,a a )=h (a ,a )a (3.26)
23 2 3 2 23 2 3 2
for all a ∈ A ,a ∈ A . Substituting a a for a in (3.23) and then subtracting
2 2 3 3 3 2 3
the right multiplication of (3.23) by a , we have
2
a [g (a ,a ),a ]=a (h (a ,a a )−h (a ,a )a ) (3.27)
3 11 1 1 2 1 13 1 3 2 13 1 3 2
for all a ∈ A ,a ∈ A and a ∈ A . Combining the relations (3.13)−(3.14),
1 1 2 2 3 3
(3.23)−(3.24) together with (3.4) yields
a h (a ,a )+a g (a ,a )=h (a ,a )a +f (a ,a )a (3.28)
1 23 2 3 3 12 1 2 13 1 3 2 12 1 2 3
for all a ∈ A ,a ∈ A and a ∈ A . Replacing a by a a in (3.28) and then
1 1 2 2 3 3 3 3 2
subtracting the right multiplication of (3.28) by a , we arrive at
2
a (h (a ,a )a −h (a ,a a ))+a [g (a ,a ),a ]
1 23 2 3 2 23 2 3 2 3 12 1 2 2
(3.29)
=(h (a ,a )a −h (a ,a a ))a
13 1 3 2 13 1 3 2 2
for all a ∈A ,a ∈A and a ∈A . Considering the identities (3.26) and (3.29),
1 1 2 2 3 3
we get
−a [g (a ,a ),a ]=(h (a ,a a )−h (a ,a )a )a (3.30)
3 12 1 2 2 13 1 3 2 13 1 3 2 2
for all a ∈A ,a ∈A and a ∈A . Making the right multiplication of (3.27) by
1 1 2 2 3 3
a and then subtracting the left multiplication of (3.30) by a , we obtain
2 1
a a [g (a ,a ),a ]=a [a ,g (a ,a )]a
1 3 12 1 2 2 3 2 11 1 1 2
10 XINFENGLIANG,ZHANKUIXIAOANDFENGWEI
for all a ∈A ,a ∈A and a ∈A . According to (3.10), we have
1 1 2 2 3 3
a [f (a ,a ),a ]a =a [a ,g (a ,a )]a
1 22 2 2 1 3 3 2 11 1 1 2
for all a ∈A ,a ∈A and a ∈A . Therefore
1 1 2 2 3 3
a [f (a ,a ),a ] 0
1 22 2 2 1 ∈Z(T).
(cid:20) 0 [a2,g11(a1,a1)]a2 (cid:21)
Commuting with b ∈ A , we get [g (a ,a ),a ][a ,b ] = 0. Then Lemma 3.2
2 2 11 1 1 2 2 2
implies g (a ,a ) ∈ Z(A ) and hence a 7→ f (a ,a ) is a commuting linear
11 1 1 2 1 12 1 2
mapping for each a ∈ A by Lemma 3.7. Similarly, we have f (a ,a ) ∈ Z(A )
2 2 22 2 2 1
and a 7→g (a ,a ) is a commuting linear mapping for each a ∈A . (cid:3)
2 12 1 2 1 1
ProofofTheorem3.4. Letq: T×T −→T beanarbitraryR-bilinearmappingof
T. It followsfromLemma 3.7,Lemma 3.9 andLemma 3.10that everycentralizing
trace of q is commuting. Then the desired result can be obtained by [8, Theorem
3.1]. (cid:3)
An algebra A overa commutative ring R is said to be central overR if Z(A)=
R1. Thefollowingtechnicallemmawillbeusedtodealwiththecentralizingtraces
of upper triangular matrix algebras.
Lemma 3.11. Let T =[RM] be a 2-torsion free triangular algebra over the com-
O B
mutative ring R and q: T ×T −→T be an R-bilinear mapping. Suppose that B is
noncommutative and both T and B are central over R. If
(1) each commuting linear mapping on B is proper,
(2) for any r ∈R and m∈M, rm=0 implies r =0 or m=0,
(3) there exist m ∈ M and b ∈ B such that m b and m are linearly inde-
0 0 0 0 0
pendent over R,
then each centralizing trace T :T −→T of q is proper.
q
Proof. We use the same notations of Theorem 3.4. Since A =R is commutative,
1
thenthe equation(3.1)showsthat[F,a ]=0 andhence [G,a ]=0. Thereforethe
1 2
centralizingtraceT is commuting. Nowthe desiredresultfollowsfrom[8, Lemma
q
3.2]. (cid:3)
Corollary 3.12. Let R be a 2-torsion free commutative domain and T (R)(n ≥
n
2) be the algebra of all n × n upper triangular matrices over R. Suppose that
q: T (R)×T (R) −→ T (R) is an R-bilinear mapping. Then every centralizing
n n n
trace T :T (R)−→T (R) of q is proper.
q n n
Proof. The proof is similar with that of [8, Corollary 3.4] and hence we omit it
here. (cid:3)
Applying Theorem 3.4 and [8, Corollary 3.5] yields
Corollary 3.13. Let H be a Hilbert space, N be a nest of H and Alg(N) be the
nest algebra associated with N. Suppose that q: Alg(N)×Alg(N) −→ Alg(N) is
an R-bilinear mapping. Then every centralizing trace T : Alg(N) −→Alg(N) of
q
q is proper.